Question

A satellite orbiting the moon very near the surface has a period of $110$min. What is free-fall acceleration on the surface of the moon?

Short answer

$g_{moon}=1.58\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given information

Given in the question that, A satellite orbiting the moon very near the surface has a period of $110$min.

Period of 

Given:

$T_{\text{sat}}=110\min$

$T_{\text{sat}}=110\min \times \frac{60\mathrm{s}}{1\min }$

$T_{\text{sat}}=6600\mathrm{s}$

Step 3:A satellite is orbiting the moon 

When a satellite is orbiting the moon at a constant speed, the only force that is acting on it is the gravitational force from the moon.

$\sum F=m{g}_{moon}=m{a}_r$

$m̸g_{\text{moon}}=m̸a_r$

$g_{\text{moon}}=\frac{v_{sat}^2}{r}$

$g_{\text{moon}}=\frac{v_{\text{sat}}^2}{1.74\times {10}^6}$ (2)

$v_{\text{sat}}=\frac{2\pi r}{T}$

$v_{\text{sat}}=\frac{2\pi \times 1.74\times {10}^6}{6600}$
$v_{sat}=1.66\times {10}^3\mathrm{m}/\mathrm{s}\left(2\right)$

From (1) and(2)

$g_{\text{moon}}=\frac{{\left[1.66\times {10}^3\right]}^2}{1.74\times {10}^6}$

$g_{moon}=1.58\mathrm{m}/{\mathrm{s}}^2$

Final answer

The free-fall acceleration on the surface of the moon is

$g_{moon}=1.58\mathrm{m}/{\mathrm{s}}^2$